Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $t = \dfrac{y - 7}{y^2 + y - 42} \div \dfrac{y + 5}{y^2 + 12y + 35} $
Dividing by an expression is the same as multiplying by its inverse. $t = \dfrac{y - 7}{y^2 + y - 42} \times \dfrac{y^2 + 12y + 35}{y + 5} $ First factor out any common factors. $t = \dfrac{y - 7}{y^2 + y - 42} \times \dfrac{y^2 + 12y + 35}{y + 5} $ Then factor the quadratic expressions. $t = \dfrac {y - 7} {(y + 7)(y - 6)} \times \dfrac {(y + 7)(y + 5)} {y + 5} $ Then multiply the two numerators and multiply the two denominators. $t = \dfrac {(y - 7) \times (y + 7)(y + 5) } { (y + 7)(y - 6) \times (y + 5)} $ $t = \dfrac {(y + 7)(y + 5)(y - 7)} {(y + 7)(y - 6)(y + 5)} $ Notice that $(y + 7)$ and $(y + 5)$ appear in both the numerator and denominator so we can cancel them. $t = \dfrac {\cancel{(y + 7)}(y + 5)(y - 7)} {\cancel{(y + 7)}(y - 6)(y + 5)} $ We are dividing by $y + 7$ , so $y + 7 \neq 0$ Therefore, $y \neq -7$ $t = \dfrac {\cancel{(y + 7)}\cancel{(y + 5)}(y - 7)} {\cancel{(y + 7)}(y - 6)\cancel{(y + 5)}} $ We are dividing by $y + 5$ , so $y + 5 \neq 0$ Therefore, $y \neq -5$ $t = \dfrac {y - 7} {y - 6} $ $ t = \dfrac{y - 7}{y - 6}; y \neq -7; y \neq -5 $